Calc Height Dielectric Oil Coaxial Cylinder Tubes

[relativespeak]

Homework Statement

Fromm Griffiths
Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of dielectric oil (susceptibility $\chi_{epsilon}$, mass density $\rho$. The inner one is maintained at potential V, and the outer one is grounded. To what height (h) does the oil rise in the space between the tubes?

So E=0 when r<a. And the potential from a to b is -V, right?

Homework Equations

$F=-\frac{dW}{dx}$
$-\frac{dW}{dx}=\frac{1}{2}V^{2}\frac{dC}{dx}$
$\frac{dC}{dx}=-\frac{\epsilon_{0} \chi_{e} \omega}{d}$
$F_{gravity}=F_{electric}$ on the dielectric oil
and $F_{electric}=F_{air}$ where the latter is the electric force in the air above the oil.

The Attempt at a Solution

There are two separate sections between the tubes, one air filled and the other oil filled.

C in the air section is $C= \frac{\epsilon_{0} A}{d} = \frac{2 \pi (b-a) (l-h)}{b-a} = 2 \pi (l-h)$

In the air, $W=\frac{1}{2} CV^{2}$.
$F=-\frac{dW}{dx}=\frac{1}{2}V^{2} -\frac{\epsilon_{0} \chi_{e} \omega}{d}$, where $\omega= 2 \pi (b-a), F=\frac{1}{2}V^{2} -\epsilon_{0} \chi_{e} 2 \pi$

I think I'm going wrong somewhere, but ideally with the capacitace and voltage you can calculate forces and then use mass density to find the height at which the forces are equivalent.

 

[mark726]

Dear Fromm Griffiths,

Thank you for your post and for your interest in this problem. I can offer some insights and suggestions for solving this problem.

First, you are correct in your understanding that the potential from a to b is -V, as the inner tube is maintained at potential V and the outer tube is grounded. This creates a potential difference between the two tubes, which leads to the formation of an electric field between them.

Next, you have correctly identified the equations that can be used to solve this problem. However, there are a few points that need clarification.

The first point is regarding the equation for electric force. The equation you have used, F = \frac{1}{2}V^{2} -\epsilon_{0} \chi_{e} 2 \pi, is actually the equation for electric potential energy, not electric force. The correct equation for electric force is F = \frac{1}{2}V^{2}\frac{dC}{dx}, which you have also listed in your post.

The second point is regarding the calculation of capacitance. In the air section, the capacitance is not equal to 2 \pi (l-h), as you have calculated. The correct equation for capacitance is C = \frac{\epsilon_{0} A}{d}, where A is the area of the plates and d is the distance between them. In this case, the area of the plates is not equal to 2 \pi (b-a) (l-h), as you have assumed. Instead, it is equal to 2 \pi (b-a) l, as the plates extend all the way to the bottom of the tank. This means that the correct equation for capacitance in the air section is C = \frac{2 \pi (b-a) l}{b-a} = 2 \pi l.

Finally, your equation for electric force in the air section is also incorrect. The correct equation is F = \frac{1}{2}V^{2}\frac{dC}{dx} = \frac{1}{2}V^{2}\frac{2 \pi l}{dx}. This is because the electric force is directly proportional to the rate of change of capacitance with distance.

To solve this problem, you can equate the electric force in the air section to the weight of the oil in the oil section. This will give you an equation that